The axiom of choice is a common set-theoretic axiom with many equivalents and consequences. This tag is for questions on where we use it in certain proofs, and how things would work without the assumption of this axiom.

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Closure in a topological product: is AC needed?

I'm working on a proof of $\prod_{\alpha\in\Lambda}\overline{A_\alpha}=\overline{\prod_{\alpha\in\Lambda}A_\alpha}$ in the product topology. This has been asked before, i.e. Closure in a product of ...
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2answers
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Proof that $(a,b)\subset\mathbb{R}$ is not countable. Does it use Axiom of Choice?

I used this proof to show $(a,b)$ is uncountable, but looking at it, I don't really see if it uses AC or not. Until recently I was thinking it does use AC (In the choice of the $a_n,b_n$), now I think ...
3
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1answer
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Does every set have choice sequences as long as the original set?

Given a set $X$, we say that $X$ has choice sequences of length $|I|$, denoted $CS(|X|,|I|)$, if for any $f:I\to{\cal P}(X)\setminus\{\emptyset\}$ there is a function $g:I\to X$ such that $g(x)\in ...
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How can choice fail in ZF?

I don't see how the Axiom of Choice can fail in ZF. By transfinite induction you can demonstrate larger and larger ordinals, using union and pairing to show the successor and limit steps, so that for ...
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A question regarding a theorem of Erdos and Hajnal

Consider the following theorem of Erdos and Hajnal: Definition. For any set $x$, a function $f$ is called ${\omega} $-Jonsson iff $f$: $^{\omega}x$ $\rightarrow$ x and whenever $y$$\subseteq$$x$ and ...
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CC for finite sets and equivalent condition

If we assume for all sequence of sets od cardinality exactly 2, there exist choice function. Can we prove Countable choice for finite sets
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Countable choice and totally bounded metric spaces

Can we prove that the following statement is equivalent to the axiom of countable choice (CC)? If every sequence in a metric space $X$ has a Cauchy subsequence, then $X$ is totally bounded. ...
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0answers
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CC and one of its equivalent condition [duplicate]

How we can prove: CC is equivalent to for all sequence(Xn) of non empty sets there exist (xn) which meets infinitely many (Xn) One side assuming CC is obvious. But what about converse. How we can ...
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1answer
30 views

Infinite set which is Dedekind finite and Weierstrass compactness

Weierstrass compactness states that each infinite set has a limit point. Why Infinite set which is Dedekind finite with discrete metric not Weierstrass compact.
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5answers
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Is axiom of choice necessary for proving that every infinite set has a countably infinite subset?

Is it possible to prove the following fact without axiom of choice ? " Every infinite set has a countably infinite subset". Can it be proved that axiom of choice is necessary here ?
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every infinite bounded subset of X has an limit point in X and CC

Related to questions and definitions used in question: Nearest point property and WeistrassB copactness of each infinite bounded set Proof of one side of theorem: every bounded infinite bounded ...
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2answers
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Does the “special continuum hypothesis” imply the axiom of choice?

In $\mathsf{ZF}$ set theory, does the "special continuum hypothesis" imply the axiom of choice, or is the axiom of choice independent of it? Here, by the "special continuum hypothesis" we mean the ...
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1answer
128 views

Can the generalized continuum hypothesis be disguised as a principle of logic?

A cool way to formulate the axiom of choice (AC) is: AC. For all sets $X$ and $Y$ and all predicates $P : X \times Y \rightarrow \rm\{True,False\}$, we have: $$(\forall x:X)(\exists y:Y)P(x,y) ...
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2answers
52 views

The generalized Axiom of Dependent Choice (DC) - is this a valid generalization?

After studying the axiom of dependent choice, I tried to think of a possible generalization of the axiom that would work in a similar way on infinite uncountable sets: by replacing the binary relation ...
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2answers
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Is there a turing machine for which halting is equivalent to the Axiom of Choice or its negation?

As seen in "A Turing machine for which halting is outside ZFC", Gödel's incompletness theorem can that there a turing machines for which halting can not be decided. My question is, is there a turing ...
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1answer
116 views

Topological properties of $[0,\omega_1)$ without choice.

Reading this wikipedia article, I arrived at the fact that $\omega_1$ can exist without choice. Since the proof I know of the fact that $[0,\omega_1)$ is sequentially compact depends on the fact that ...
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1answer
64 views

Why does a proof of $\exists f: X\to Y$ injection $\iff \exists g: Y \to X$ surjection requires the axiom of choice?

Why does a proof of $\exists f: X\to Y$ injection $\iff \exists g: Y \to X$ surjection requires the axiom of choice? This question is answered here: There exists an injection from $X$ to $Y$ if and ...
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1answer
30 views

Can it be proved without the axiom of choice that every cardinal is comparable with every finite cardinal?

Can it be proven in ZF, without using the axiom of choice, that every finite set is a universal size comparator, meaning, is comparable with every set in terms of size? And what is the proof?
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Over ZF does “every non-seperable Hilbert space has an orthonormal basis” imply “there exists a non-Lebesgue measurable set”?

I know from this question that it's an open problem whether or not the existence of a dense orthonomral basis for every real or complex Hilbert space $(\text{B}_\text{orth})$ implies the axiom of ...
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2answers
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Does construction of infinite product measure require axiom of choice?

I am learning about infinite (countable) product measure, which the exact statement of the theorem I write below. I was wondering if the theorem requires axiom of choice or not. I would appreciate any ...
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Indexed sum of cardinals [duplicate]

Let $\{ \kappa_i | i\in I \}$ be an indexed set of cardinals. We define the sum as: $\sum\limits_{i\in I}\kappa_i = \left\vert \bigcup\limits_{i\in I}X_i \right\vert$, where $\left\vert X_i ...
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1answer
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Approximating nice functions with wild ones

Let $X$ and $Y$ be toplogical spaces, and call a function $f:X\to Y$ wild if the preimage $f^{-1}(\{y\})$ is dense in $X$ for every $y\in Y$ -- or, equivalently, if the image of every nonempty open ...
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2answers
120 views

Explicit construction of a nonmeasurable set, where only the proof of correctness uses choice?

By Solovay's theorem, assuming the existence of an inaccessible cardinal, the axiom of choice is necessary to prove the existence of nonmeasurable sets. In the past, I've thought that one consequence ...
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1answer
71 views

Does Lowenheim-Skolem theorem depend on axiom of choice?

The proofs of Lowenheim-Skolem I have seen all depended on the use of choice functions. Is there any proof not dependent on axiom of choice? Or is Lowenheim-Skolem a result of axiom of choice?
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1answer
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Find a uniquely defined element in this $\aleph_1$-indexed Cartesian product

Denote by $A$ the set of all ordinals with cardinality exactly equal to ${\aleph}_0$, and for $\alpha\in A$ let $B_{\alpha}$ denote the set of all bijections between $\alpha$ and $\omega$ ; finally ...
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3answers
497 views

Does this proposition from complex analysis depend on AC?

I was reading III vol. of Princeton lectures on analysis. Proposition 1.4: "If $\Omega_{1}\supset\Omega_{2}\supset\ldots\supset\Omega_{n}\supset\ldots $ is a sequence of non-empty compact sets in ...
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123 views

Where is the axiom of choice used in Rudin's proof of “the countable union of countable sets is countable”?

Baby Rudin proves that the countable union of countable sets is countable. From reading other proofs online, the axiom of choice has to be invoked; however, I'm not seeing immediately where that is ...
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858 views

W. Mückenheim claims a severe inconsistency of transfinite set theory; true? [closed]

The abstract for a paper on arxiv.org (http://arxiv.org/pdf/math/0408089v3.pdf) reads (with my emphasis): "Transfinite set theory including the axiom of choice supplies the following basic theorems: ...
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1answer
62 views

If a set $S$ has a choice function, does $\bigcup S$ have one too?

I have an exercise in a book that asserts that if a set $S$ has a choice function on it, then so does the union of all its elements $\bigcup S$ (without assuming the axiom of choice). I, however, have ...
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2answers
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If $n < \aleph^*(m)$, then $n < 2^m$.

Without $AC$ Let $\aleph^*(m)$ be the least aleph that $\not\leq^* m$. I need a help or hint that if $n < \aleph^*(m)$, then $n < 2^m$. $a \leq^* b$ means we can define a surjective map from ...
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3answers
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Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice)

I am interested in proving the titular claim: Transitive Closure of a Well-Founded Relation is Well-Founded (without Axiom of Choice) My approach: Let $R$ be a well-founded relation. We ...
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2answers
368 views

Does the definition of the fundamental group implicitly assume the Axiom of Choice?

Okay, I'm a little foggy around the axiom of choice, so help me out here. The standard way the fundamental group of a connected space $X$ is defined is as follows. You consider the set of all loops ...
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2answers
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Good resources for studying independence proofs

I've finished most of Enderton's set theory. And I intend to spend some time studying independence proofs. I'm more interested in independence of axiom of choice not CH. From I know so far, there ...
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1answer
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A Banach space of (Hamel) dimension $\kappa$ exists if and only if $\kappa^{\aleph_0}=\kappa$

A Banach space of (Hamel) dimension $\kappa$ exists if and only if $\kappa^{\aleph_0}=\kappa$. How will we prove the converse implication. One sided implication for Hilbert Space is proved in ...
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1answer
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What is the significance of having Prime Ideal Theorem in models for failure of Axiom of Choice?

Prime Ideal Theorem says: PIT: Every ideal on a Boolean algebra can be extended to a prime ideal. It follows from Axiom of Choice but is weaker than it. In many cases I saw that people check ...
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1answer
58 views

Basis for $\mathbb{R}^{\infty}$ [duplicate]

It follows from Zorn's Lemma that every vector space $V$ has a basis (this means, a subset $B$ of $V$ that generate any $v \in V$ by means a finite linear combination, and such that $B$ is LI) . But, ...
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1answer
23 views

The least $\aleph$ that has no surjective map from $m$ to it.

Without $AC$. Let $\aleph^*(m)$ be the least aleph that $\not\leq^* m$. How to show that $\aleph^*(m)$ exists and $\aleph^*(m)= \{\alpha\in ON\mid\ \alpha\leq^*m\}$. $ON$ is the class of all ...
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0answers
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onto map implies existence of one one map and AC [duplicate]

Let us assume the fact: $f: A \to B$ onto function implies there exist $1-1$ function from $B$ to $A$. Would it imply AC? I know every surjective function has right inverse this fact is ...
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1answer
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dual Dedekind-infinity may not imply Dedekind-infinite without AC

It is written in wikipedia: https://en.wikipedia.org/wiki/Dedekind-infinite_set It is not provable (in ZF without the AC) that dual Dedekind-infinity implies that A is Dedekind-infinite. (For ...
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2answers
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Does the principle of schematic dependent choice follow from ZFCU?

Let ZFCU be ZFC modified in the usual way to allow for urelements but without an axiom stating that there is a set of all urelements. Let the principle of Schematic Dependent Choice (SDC) be: ...
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1answer
72 views

Are there weak versions of the axiom of choice equivalent to weak versions of Zorn's lemma and similar principles?

I recalled reading about other weaker forms of $AC$, for example countable choice, where we could make choices from a sequence $(S_{k})_{k \in \mathbb{N}}$ of non-empty sets. I also recalled mention ...
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Does there exist a model of $ZF¬C$ in which there is a function on $\mathbb R$ which is sequentially continuous at a point where it is not continuous? [duplicate]

Does there exist a model of $ZF¬C$ in which there is a function $f:\mathbb R \to \mathbb R$ such that $f$ is sequentially continuous at some $a \in \mathbb R$ but not $\epsilon-\delta$ continuous , ...
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“Sequential continuity is equivalent to $\epsilon$-$\delta $ continuity ” implies Axiom of countable choice for collection of subsets of $\mathbb R$?

"A function $f: \mathbb R \to \mathbb R$ is continuous at $x \in \mathbb R$ , if and only if it is sequentially continuous " , does this statement imply "the Axiom of Choice for countable collections ...
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1answer
65 views

Proof that every field $F$ has an algebraic closure $\bar F$

I am reading the book A First Course in Abstract Algebra written by Fraleigh and I do not really understand the proof of theorem 31.22, that every field $F$ has and algebraic closure $\bar F$. I ...
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1answer
75 views

First Uncountable Ordinal Cofinality: Needs AC?

Say $\omega_1$ is the first uncountable ordinal. The reason I care about $\omega_1$ is Any countable subset of $\omega_1$ is bounded (or if you prefer, there is no countable cofinal subset). This ...
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(Non-Hopfian) groups that only have quotients that are themselves or the trivial group.

A group is non-Hopfian provided it is isomorphic to a proper quotient. The classic, finitely presented, example of such a group is the Baumslag-Solitar group $$BS(2,3)= \langle x,t \mid t^{-1}x^2 t ...
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1answer
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ZF and the Existence of Finitely additive measure on $\mathcal{P}(\mathbb{R})$

My understanding is that Solovay (1970)'s relative consistency shows that if ZFC+I has a model then ZF+DC has a model in which every subset of the reals is Lebesgue measurable (and hence ...
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133 views

The non-existence of non-principal ultrafilters in ZF

In Hrbacek and Jech (1999, p.205), they point out that "it is known that the theorem [the extension of any filter to an ultrafilter] cannot be proved in Zermelo-Fraenkel set theory alone." And in Jech ...
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2answers
678 views

Do we know that we can't define a well-ordering of the reals?

Folklore has it that it is impossible to define a well-ordering of the reals explicitly. There exist pointwise definable models of ZFC where every set is definable without parameters: it is the ...
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undecidable statement of the form “$F$ is a choice function on $M$”

Are there unary predicates $\varphi(x), \psi(x)$ such that The formula that states "There is a set $M$ with: $\forall x[x\in M \leftrightarrow \varphi(x)]$" is provable in $ZFC$. The formula that ...